Chapter 7

[Pages:25]Chapter 7

Exercise 7A

1. I will use the "intelligent guess" method for this

question, but my preference is for the "rearrang-

ing" method, so I will use that for most of the

questions where one of these approaches is suit-

able.

Guess

y

=

(2x+5)5 5

d (2x+5)5 dx 5

=

(2x + 5)4(2)

Adjust

by

a

factor

of

1 2

:

(2x + 5)5 +c

10

2. dy = (3x + 1)3 dx

=

1 (3(3x +

1)3)

3

1 (3x + 1)4

y=

+c

34

(3x + 1)4

=

+c

12

3. This can not be rearranged to the form f (x)(f(x))n so we expand:

dy = x(3x + 4)

dx = 3x2 + 4x

3x3 4x2 y= + +c

32 = x3 + 2x2 + c

4. dy = 50(1 + 5x)4 dx = 10(5(1 + 5x)4)

(1 + 5x)5

y = 10

+c

5

= 2(1 + 5x)5 + c

5. dy = 24x(2 - x2)3 dx = -12(-2x(2 - x2)3)

(2 - x2)4

y = -12

+c

4

= -3(2 - x2)4 + c

6. This can not be rearranged to the form f (x)(f(x))n so we expand:

dy = x(1 + 4x)2 dx

= x(1 + 8x + 16x2)

= x + 8x2 + 16x3

x2 8x3 16x4

y= + +

+c

23

4

= x2 + 8x3 + 4x4 + c 23

7. dy = 30x(x2 - 3)2 dx = 15(2x(x2 - 3)2)

(x2 - 3)3

y = 15

+c

3

= 5(x2 - 3)3 + c

8. This can not be rearranged to the form f (x)(f(x))n so we expand:

dy = x(2x - 3)2 dx

= x(4x2 - 12x + 9)

= 4x3 - 12x2 + 9x

4x4 12x3 9x2

y= -

+ +c

4

3

2

= x4 - 4x3 + 9x2 + c 2

9. dy = 12(2x + 1)3 dx = 6(2(2x + 1)3)

(2x + 1)4

y=6

+c

4

3(2x + 1)4

=

+c

2

10. dy = 2(3x + 1)4 dx

=

2 (3(3x

+

1)4)

3

2 (3x + 1)5

y=

+c

3

5

2(3x + 1)5

=

+c

15

11. dy = (2x - 3)4 dx

=

1 (2(2x

-

3)4)

2

1 (2x - 3)5

y=

+c

2

5

(2x - 3)5

=

+c

10

12. dy = 5x2(3 - x3)4 dx

= 5 (-3x2(3 - x3)4) -3

5 (3 - x3)5

y=-

+c

3

5

(3 - x3)5

=-

+c

3

(x3 - 3)5

=

+c

3

(Note that the last step is valid because (-1)5 =

1

Exercise 7A

Solutions to A.J. Sadler's

-1. Such a simplification would not be possible if we were raising to an even power.)

13. dy = (1 + x)4 dx

(1 + x)5

y=

+c

5

14. dy = 4x(2 + x2)4 dx = 2(2x(2 + x2)4)

(2 + x2)5

y=2

+c

5

2(2 + x2)5

=

+c

5

15. dy = (1 + 2x)4 dx

=

1 (2(1

+ 2x)4)

2

1 (1 + 2x)5

y=

+c

2

5

(1 + 2x)5

=

+c

10

16. dy = 4x(1 + x2) dx = 2(2x(1 + x2))

(1 + x2)2

y=2

+c

2

= (1 + x2)2 + c This question is probably just as easy to do by first expanding:

dy = 4x(1 + x2) dx

= 4x + 4x3

x2 x4 y=4 +4 +c

23 = 2x2 + x4 + c

Although these two solutions may not look quite the same, you should satisfy yourself that they are, in fact, both correct. (Remember, c is an arbitrary constant.)

17. dy = 60(2x - 3)5 dx = 30(2(2x - 3)5

(2x - 3)6

y = 30

+c

6

= 5(2x - 3)6 + c

18. dy = 60(3 - 2x)5 dx = -30(-2(3 - 2x)5

(3 - 2x)6

y = -30

+c

6

= -5(3 - 2x)6 + c

dy

1

19. dx = (x + 2)4

= (x + 2)-4

(x + 2)-3

y=

+c

-3

1

=

- 3(x

+

2)3

+

c

dy

1

20. dx = (2x + 1)4

= (2x + 1)-4

=

1 (2(2x

+

1)-4

2

1 (2x + 1)-3

y=

+c

2 -3

1

=

- 6(2x

+

1)3

+

c

dy

25x

21. dx = - (x2 + 1)5

= -25x(x2 + 1)-5

= -25 (2x(x2 + 1)-5) 2

-25 (x2 + 1)-4

y=

+c

2 -4

25 = 8(x2 + 1)4 + c

dy

6

22. dx = (3x + 5)3

= 6(3x + 5)-3

= 2(3(3x + 5)-3)

(3x + 5)-2

y=2

+c

-2

1

=

- (3x

+

5)2

+

c

dy

18x

23. dx = (3x2 + 5)3

= 3(6x)(3x2 + 5)-3

(3x2 + 5)-2

y=3

+c

-2

3 = - 2(3x2 + 5)2 + c

24.

dy

= 12 3 3x - 2

dx

1

= 4 3(3x - 2) 3

(3x

-

2)

4 3

y=4

4

+c

3

43

= 4(3x - 2) 3

+c

4

4

= 3(3x - 2) 3 + c

2

Unit 3C Specialist Mathematics

dy

25. = 12 3x + 5

dx

1

= 4 3(3x + 5) 2

(3x

+

5)

3 2

y=4

3

+c

2

32

= 4(3x + 5) 2

+c

3

8(3x

+

5)

3 2

=

+c

3

dy

12

26. =

dx 3x + 5

=4

3(3x

+

5)-

1 2

(3x

+

5)

1 2

y=4

1

+c

2

1

= 8(3x + 5) 2 + c

= 8 3x + 5 + c

27. dy = 3 - 12(-3x2(1 - x3)2) dx

(1 - x3)3

y = 3x - 12

+c

3

= 3x - 4(1 - x3)3 + c

28. dP = 12(3(2 + 3t)3) dt

(2 + 3t)4

P = 12

+c

4

= 3(2 + 3t)4 + c

50 = 3(2 + 3(0))4 + c

50 = 3(24) + c

c = 50 - 3 ? 16

=2

P = 3(2 + 3t)4 + 2

29. dP = 12(2t(t2 - 5)3 dt

(t2 - 5)4

P = 12

+c

4

= 3(t2 - 5)4 + c

10 = 3(22 - 5)4 + c

10 = 3(-1)4 + c

10 = 3 + c

c=7

P = 3(t2 - 5)4 + 7

Exercise 7B

Exercise 7B

1. 5 cos x dx = 5 sin x + c

2. 2 sin x dx = -2 cos x + c

3. -10 sin x dx = 10 cos x + c

4. -2 cos x dx = -2 sin x + c

5. 6 cos 2x dx = 3 2 cos 2x dx

= 3 sin 2x + c

1 6. 2 cos 6x dx = 6 cos 6x dx

3

sin 6x

=

+c

3

7. 12 sin 4x dx = -3 -4 sin 4x dx = -3 cos 4x + c

8. 9. 10. 11. 12. 3

1 - sin 3x dx = -3 sin 3x dx

3

cos 3x

=

+c

3

4 -8 cos 10x dx = - 10 cos 10x dx

5

4 sin 10x

=-

+c

5

x

1x

sin dx = -2 - sin dx

2

22

x = -2 cos + c

2

3x

2 3 3x

cos dx =

cos dx

2

32 2

2 3x = sin + c

32

2x

3

-6 sin dx = 6 ?

3

2

2 2x - sin dx

33

2x = 9 cos + c

3

Exercise 7B

Solutions to A.J. Sadler's

13. For this question and the next, you should note

that sin

x

+

2

= cos x and cos

x

-

2

= sin x.

You may make the substitution at the beginning

and then antidifferentiate, or first antidifferenti-

ate and then substitute.

14. See previous question.

2 15. cos 2x + dx

3

1

2

= 2 cos 2x +

dx

2

3

2 = sin 2x + + c

3

16. sin(-x) dx = - sin x dx

= cos x + c

4 17. cos2 x dx = 4 tan x + c

1

1

4

18. cos2 4x dx = 4 cos2 4x dx

1 = tan(4x) + c

4

1 19. cos2(x - 1) dx = tan(x - 1) + c

20. 6 cos2x + 6 sin 3x dx

= 3 2 cos 2x dx - 2 -3 sin 3x dx = 3 sin 2x - 2 cos 3x + c

21. cos8x - 4 sin 2x dx 1

= 8 cos 8x dx + 2 -2 sin 2x dx 8 1

= sin 8x + 2 cos 2x + c 8

22. 2x + 4 cos x + 6 cos 2x dx

= 2x dx + 4 cos x dx + 3 2 cos 2x dx

= x2 + 4 sin x + 3 sin 2x + c 23. Although this looks long, you should by now be

able to antidifferentiate it in a single step, simply working term by term.

3 + 4x - 6x2 dx = 3x + 2x2 - 2x3 + c1

so the overall antiderivative is 3x + 2x2 - 2x3 + 2 sin 5x + cos 4x + c 2

24.

cos3

x

sin

x

dx

=

cos4 -

x

+

c

4

25.

30

cos5

x

sin

x

dx

=

30 -

cos6

x

+

c

6

= -5 cos6 x + c

26. sin4 x cos x dx = sin5 x + c 5

27. 6 sin3 x cos x dx = 6 sin4 x + c 4

3 sin4 x

=

+c

2

28. -2 cos4 x sin x dx = 2 cos5 x + c 5

29.

-2

sin7

x

cos

x

dx

=

2 -

sin8

x

+

c

8

sin8 x

=-

+c

4

30. 32 sin3 4x cos 4x dx = 8 sin3 4x(4 cos 4x) dx

8 sin4 4x

=

+c

4

= 2 sin4 4x + c

31. -24 sin3 2x cos2x dx

= -12 sin3 2x(2 cos 2x) dx

-12 sin4 2x

=

+c

4

= -3 sin4 2x + c

32. 20 sin4 2x cos 2x dx

= 10 sin4 2x(2 cos 2x) dx

10 sin5 2x

=

+c

5

= 2 sin5 2x + c

10 cos 5x dx = 2 5 cos 5x dx

= 2 sin 5x + c2 1

-2 sin 4x dx = -4 sin 4x dx 2 cos 4x

= 2 + c3

33. 4

-6 cos2 4x sin4x dx

3 =

cos2 4x(-4 sin 4x) dx

2

3 cos3 4x

=

+c

2?3

cos3 4x

=

+c

2

Unit 3C Specialist Mathematics

Exercise 7B

34. sin3 x dx = sin x(sin2 x) dx

= sin x(1 - cos2 x) dx

= sin x - sin x cos2 x dx

cos3 x

= - cos x +

+c

3

35. cos3 x dx = cos x(cos2 x) dx

= cos x(1 - sin2 x) dx

= cos x - cos x sin2 x dx

sin3 x

= sin x -

+c

3

39. 8 sin4 x dx = 8 (sin2 x)2 dx

-2 sin2 x 2

=8

dx

-2

(-2 sin2 x)2

=8

dx

4

= 2 (-2 sin2 x)2 dx

= 2 (1 - 2 sin2 x - 1)2 dx

= 2 (cos 2x - 1)2 dx

= 2 cos2 2x - 2 cos 2x + 1 dx

= 2 cos2 2x - 4 cos 2x + 2 dx

= 2 cos2 2x - 1 - 4 cos 2x + 3 dx

36. cos5x dx = cos x(cos4 x) dx = cos x(cos2 x)2 dx

= cos 4x - 4 cos 2x + 3 dx

sin 4x 4 sin 2x

=

-

+ 3x + c

4

2

sin 4x

=

- 2 sin 2x + 3x + c

4

= cos x(1 - sin2 x)2 dx

40. cos2 x + sin2 x dx = 1 dx

= cos x(1 - 2 sin2 x + sin4 x) dx = cos x - 2 cos x sin2 x + cos x sin4 x dx

=x+c (Compare this with your answers for questions 37 and 38.)

2 sin3 x sin5 x

= sin x -

+

+c

3

5

37. cos2 x dx = 1 2 cos2 x dx 2

1 =

2 cos2 x - 1 + 1 dx

2

1 = cos 2x + 1 dx

2

1 sin 2x

=

+x +c

22

sin 2x x

=

+ +c

42

38. sin2 x dx = - 1 -2 sin2 x dx 2

1 =-

1 - 2 sin2 x - 1 dx

2

1 = - cos 2x - 1 dx

2

1 sin 2x

=-

-x +c

22

sin 2x x

=-

+ +c

42

41. cos2 x - sin2 x dx = cos 2x dx

sin 2x

=

+c

2

(Compare this with your answers for questions

37 and 38.)

42. sin 5x cos 2x + cos 5x sin 2x dx

= sin(5x + 2x) dx

= sin 7x dx

cos 7x

=-

+c

7

(You need to know the angle sum trig identities

well enough to recognise them.)

43. sin 3x cos x - cos 3x sin x dx

= sin(3x - x) dx

= sin 2x dx

cos 2x

=-

+c

2

5

Exercise 7B

Solutions to A.J. Sadler's

44. cos 5x cos 2x - sin 5x sin 2x dx

= cos(5x + 2x) dx

= cos 7x dx

sin 7x

=

+c

7

45. cos 5x cos x + sin 5x sin x dx

= cos(5x - x) dx

= cos 4x dx

sin 4x

=

+c

4

46. Refer to your answers to questions 34 and 37.

47. There are (at least) three different ways of approaching this question. You can treat it as

? 2f (x)f (x) dx = [f (x)]2 + c where f (x) = sin x (i.e. sin2x + c);

? -2f (x)f (x) dx = - [f (x)]2 + c where f (x) = cos x (i.e. -cos2x + c); or

? Recognise the double angle formula for sine:

sin

2x

dx

=

-

1 2

cos

2x

+

c.

You should satisfy yourself that these answers are equivalent (differing only in the value of the constant of integration).

48. sin3 x cos2 x dx

= sin x sin2 x cos2 x dx

= sin x(1 - cos2 x) cos2 x dx

= sin x cos2 x - sin x cos4 x dx

cos3 x cos5 x

=-

+

+c

3

5

49. cos3 x sin2 x dx

= cos x cos2 x sin2 x dx

= cos x(1 - sin2 x) sin2 x dx

= cos x sin2 x - cos x sin4 x dx

sin3 x sin5 x

=

-

+c

3

5

50.

C = 4 sin 2p dp

= -2 cos 2p + k 3 = -2 cos 0 + k k=5 C = 5 - 2 cos 2p

Note that I use k for the constant of integration here rather than the more commonly used c so as to avoid confusion with C.

51.

P = 12 cos 3x dx

= 4 sin 3x + c 3

10 = 4 sin + c 2

10 = -4 + c c = 14

P = 4 sin 3x + 14

52. 2 sec x tan x dx = 2 sec x + c

53. By the chain rule,

d (sec 2x) = 2 sec 2x tan 2x

dx so

sec 2x

sec 2x tan 2x dx =

+c

2

54. By the chain rule,

d (cot 2x) = -2 cosec2 2x dx so

-4 cosec2 2x dx = 2 cot 2x + c

55.

1 sin2 x dx =

cosec2 x dx

= - cot x + c

sin x

1 sin x

56.

cos2 x dx =

dx cos x cosx

= sec x tan x dx

= sec x + c

cos x 57. sin2 x dx = cosec x cot x dx

= - cosec x + c

58. This is of the form f (x)nf (x) dx where f (x) = cosec x:

20(cosec x cot x)(cosec3 x) dx = -20 cosec4 x + c 4

= -5 cosec4 x + c

59. 20 cosec4 x cot x dx = 20(cosec x cot x)(cosec3 x) dx = -5 cosec4 x + c

6

Unit 3C Specialist Mathematics

Exercise 7C

In the following solutions I have used notation like du = 2x dx. This is convenient in helping us make the variable substitutions, but it is important to remember that du and dx are meaningless outside the context of integration. We must not fall into the trap of treating them like real quantities.

1. u = x2 - 3 du = 2x dx

60x(x2 - 3)5 dx = 30 u5 2x dx = 30 u5 du = 5u6 + c = 5(x2 - 3)6 + c

Exercise 7C

12x(3x + 1)5 dx

= 12 u - 1 u5 1 du

3

3

= 4 (u - 1)u5 du 3

4 =

u6 - u5 du

3

4 u7 u6

=

- +c

37 6

4u7 2u6 = - +c

21 9

2u6 = (6u - 7) + c

63

2(3x + 1)6

=

(6(3x + 1) - 7) + c

63

2(3x + 1)6

=

(18x + 6 - 7) + c

63

= 2 (3x + 1)6(18x - 1) + c 63

2. u = 1 - 2x du = -2 dx

1-u

1

x=

dx = - du

2

2

80x(1 - 2x)3 dx

=

80

1-u

u3

1 -

du

2

2

= -20(1 - u)u3 du

= 20 u4 - u3 du

u5 u4 = 20 - + c

54 = 4u5 - 5u4 + c = u4(4u - 5) + c = (1 - 2x)4 (4(1 - 2x) - 5) + c = (1 - 2x)4(4 - 8x - 5) + c = -(1 - 2x)4(8x + 1) + c

4. u = 2x2 - 1 du = 4x dx

6x(2x2 - 1)5 dx = 3 u54x dx 2

3 =

u5 du

2

3 u6 = +c

26

= 1 u6 + c 4

= 1 (2x2 - 1)6 + c 4

5. u = 3x2 + 1 du = 6x dx

12x(3x2 + 1)5 dx = 2u56x dx

= 2 u5 du u6

=2 +c 6

= 1 u6 + c 3

= 1 (3x2 + 1)6 + c 3

3. u = 3x + 1 du = 3 dx

u-1

1

x=

dx = du

3

3

6. u = x - 2 du = dx x = u + 2 dx = du

7

Exercise 7C

Solutions to A.J. Sadler's

3x(x - 2)5 dx = 3(u + 2)u5 du

= 3u6 + 6u5 du

u7 u6 =3 +6 +c

76

=

u7 3

+

u6

+

c

7

= 1 u6(3u + 7) + c 7

= 1 (x - 2)6 (3(x - 2) + 7) + c 7

= 1 (x - 2)6(3x - 6 + 7) + c 7

= 1 (x - 2)6(3x + 1) + c 7

7. u = 3 - x du = -dx x = 3 - u dx = -du

20x(3 - x)3 dx = -20(3 - u)u3 du

9. u = 2x + 3 du = 2 dx

u-3

1

x=

dx = du

2

2

20x(2x + 3)3 dx

= 5(u - 3)u3 du

= u5 - 15u4 + c 4

= 1 u4(4u - 15) + c 4

= 1 (2x + 3)4 (4(2x + 3) - 15) + c 4

= 1 (2x + 3)4(8x - 3) + c 4

10. u = 3x + 1 du = 3 dx

u-1

1

x=

dx = du

3

3

= 20 u4 - 3u3 du

u5 3u4

= 20 -

+c

54

= 4u5 - 15u4 + c

= u4(4u - 15) + c

= (3 - x)4 (4(3 - x) - 15) + c

= (3 - x)4(12 - 4x - 15) + c

= (3 - x)4(-4x - 3) + c

= -(3 - x)4(4x + 3) + c

8. u = 5 - 2x du = -2 dx

5-u

1

x=

dx = - du

2

2

18x 3x + 1 dx

1

= 2(u - 1)u 2 du

3

1

= 2 u 2 - u 2 du

2u

5 2

2u

3 2

=2

-

+c

5

3

=

4

u3(3u

-

5)

+

c

15

4 =

(3x + 1)3 (3(3x + 1) - 5) + c

15

4 =

(3x + 1)3(9x - 2) + c

15

4x(5 - 2x)5 dx

=

4 5 - u u5

1 -

du

2

2

= (u - 5)u5 du

11. u = 3x2 + 5 du = 6x dx

u7 5u6 = - +c

76 = 1 u6(6u - 35) + c

42 = 1 (5 - 2x)6 (6(5 - 2x) - 35) + c

42 = 1 (5 - 2x)6(-12x - 5) + c

42 = - 1 (5 - 2x)6(12x + 5) + c

42

6x dx

du

=

3x2 + 5

u

=2 u+c

= 2 3x2 + 5 + c

12. u = 1 - 2x du = -2 dx

1-u

1

x=

dx = - du

2

2

8

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